I have problem with solving following equation with initial conditions: $$y*y''-2(y')^2-y^2=0 $$ $$y(0)=1; y'(0)=0 $$ The problem is that i've tried substitution $ u(y)=y' $ and I end up with $$u'*u-2u^2/y =y $$ which is basically bernouli equation. I've done z sub so that: $z(y)=u^2$ and got equation $$z'-4z/y=2y $$ I solved that and got (with my initial condition ) $z=y^4-y^2 $ and that implies $ u^2=y^4-y^2 $ I have no idea what should be next step Any help appreciated !
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it does not depend on 3 variables – paweta Jun 21 '20 at 18:41
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both equations after u sub are dependend on y – paweta Jun 21 '20 at 18:42
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but y is a function of x – paweta Jun 21 '20 at 18:42
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it is u[y(x)] . – paweta Jun 21 '20 at 18:45
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look at my profile, i asked simmilar question and somebody solved it using u=y' sub – paweta Jun 21 '20 at 18:46
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u can use u(y)=y' sub and u get that y''=u'y'=u'u,( derivative of a compound function) u can use also u(x)=y but not in this case – paweta Jun 21 '20 at 18:49
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HINT
To begin with, notice that \begin{align*} yy'' - 2(y')^{2} - y^{2} = 0 \Longleftrightarrow \frac{y''}{y} - \frac{(y')^{2}}{y^{2}} - \left(\frac{y'}{y}\right)^{2} - 1 = 0 \end{align*} Moreover, we do also have that \begin{align*} \frac{y''}{y} - \frac{(y')^{2}}{y^{2}} = \frac{y''y - (y')^{2}}{y^{2}} = \left(\frac{y'}{y}\right)' \end{align*}
Hence, if we make the substitution $y' = uy$, we obtain the following ODE \begin{align*} u' - u^{2} - 1 = 0 \Longleftrightarrow u' = u^{2} + 1 \Longleftrightarrow \frac{u'}{u^{2}+1} = 1 \end{align*}
Can you take it from here?
user0102
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